Did you try it out? It just works(tm).

Er... yes. It works exactly how you'd expect it to. Or, at least, how I'd expect it to. What did you try? What happened when you tried it?

sub a_subroutine {
  print "this is a subroutine\n";
}

sub another_subroutine {
  print "this is another subroutine\n";
  a_subroutine();
}

another_subroutine();

I think you should spend some time in the first chapters of some Perl books for beginners, like Learning Perl by our own merlyn, or Beginning Perl by Simon Cozens. If you look closely at the latter page, you can find a link to a free online version.

use strict;
use warnings;

bongo(\&bingo);

sub bingo {
    print "bingo";
}

sub bongo {
    my $p = $_[0];
    $p->();
}

Re^2: To call a sub from another sub

Anonymous Monk

created: 2006-02-02 06:55:04

perlmonks

• reply
• node
• thread
• author

• node
• author
• thread

prlmnks

• node
• thread
• author

I would like that subrutine1 sbould get a value from subrutine2, It should process this value and return the result to subrutine2. Is this possible or do I have to pass subrutine1 to subrutine2?

Re^3: To call a sub from another sub

jkva

created: 2006-02-02 07:02:36

perlmonks

• reply
• node
• thread
• author

• node
• author
• thread

prlmnks

• node
• thread
• author

Perhaps you want this :

sub subroutine_1 {
  my $val = subroutine_2();
  print "$val this is subroutine_2 output!";
}

sub subroutine_2 {
  return "w00t!";
}

Re: To call a sub from another sub

jonadab

created: 2006-02-02 07:26:43

perlmonks

• reply
• node
• thread
• author

• node
• author
• thread

prlmnks

• node
• thread
• author

Two points:

• You missed the part where a value is passed in, processed, and returned.
• Don't just hand him complete working code. He made no attempt to write any code himself, and he won't learn anything until he does, and it's obviously homework, the first assignment of the semester by the look of it.

Re: To call a sub from another sub

jonadab

created: 2006-02-02 07:19:22

perlmonks

• reply
• node
• thread
• author

• node
• author
• thread

prlmnks

• node
• thread
• author

Yes, it'll work just fine. It will even work if the outer subroutine is a lexical closure and passes the other subroutine one of the lexically-retained references it is holding, like this:

sub :: {return lc pop}
sub jonadab {
   my $lex = shift;
   @ARGV=qw(http://);
   return sub{
     my ($ical) = @_;
     my $closure = :: $lex.".".$ical."/".$/;
}}
my $foo = jonadab("PerlMonks");
$\=$foo->("Org");
print shift;

Note: if you turn in code like that for homework, don't expect a good grade.

use strict;
use warnings;

whatever();

sub whatever {
    my $value = 5;
    $value = increment($value);
    print $value;
}

sub increment {
    return $_[0] + 1;
}